Linear depedency

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Prove whethere is linearly independent or not the following:

{[tex]e^x,e^{2x}[/tex]}
 

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  • #2
jbunniii
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If not, then there must exist two scalars, a and b (they can't both be zero), such that

[tex]a e^x + b e^{2x} = 0[/tex]

for all x. Can you find such scalars?
 
  • #3
HallsofIvy
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Since that is to be true for all x, you can get two simple equations to solve for a and b by choosing two values of x.
 
  • #4
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I think it's possible:
[tex]a e^x + b e^{2x} = 0[/tex] <==> [tex]a e^x = - b e^{2x}[/tex] <==> [tex]ln (a e^x) = - ln (b e^{2x})[/tex] <==> ax = -2bx

So if we carefully choose a, we can definitely show your given set is linearly dependent.
 
  • #5
jbunniii
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I think it's possible:
[tex]a e^x + b e^{2x} = 0[/tex] <==> [tex]a e^x = - b e^{2x}[/tex] <==> [tex]ln (a e^x) = - ln (b e^{2x})[/tex] <==> ax = -2bx

So above, if we carefully choose a, we can definitely show your given set is linearly dependent.
Except

[tex]\ln(a e^x) = \ln(a) + x[/tex]
[tex]\ln(-b e^{2x}) = \ln(-b) + 2x[/tex]

so you need

[tex]\ln(a) - \ln(-b) = x[/tex] for all x!
 
  • #6
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Except

[tex]\ln(a e^x) = \ln(a) + x[/tex]
[tex]\ln(-b e^{2x}) = \ln(-b) + 2x[/tex]

so you need

[tex]\ln(a) - \ln(-b) = x[/tex] for all x!
Bummer, so sorry.
 
  • #7
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If not, then there must exist two scalars, a and b (they can't both be zero), such that

[tex]a e^x + b e^{2x} = 0[/tex]

for all x. Can you find such scalars?
Put x=0,then a+b=0 ====> a=-b

Hence the above are linearly dependent

CORRECT???

THANKS
 
  • #8
matt grime
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No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.
 
  • #9
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No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.

Mat grime ,please ,write down for me the definition ,when a set of n vectors are linearly independent
 
  • #10
matt grime
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Apologies, I missed out an 'in'. The functions e^x and e^2x are clearly linearly independent (over R), which is where I misunderstood what you're saying. (If you don't see it then just rearrange ae^x +be^2x=0 to see that you're claim of linear dependence implies that e^x = -a/b for all x.)
 
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  • #11
HallsofIvy
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Put x=0,then a+b=0 ====> a=-b

Hence the above are linearly dependent

CORRECT???

THANKS
No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.

Another way to prove it is this: if [itex]ae^x+ be^{2x}= 0[/itex] for all x, then it is a constant and its derivative must be 0 for all x: [itex]ae^x+ 2be^{2x}= 0[/itex]. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.
 
  • #12
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No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.

Another way to prove it is this: if [itex]ae^x+ be^{2x}= 0[/itex] for all x, then it is a constant and its derivative must be 0 for all x: [itex]ae^x+ 2be^{2x}= 0[/itex]. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.
Is not the definition for linear indepedence the following:

for all a,b and x ,real Nos and [tex]ae^x + be^{2x}=0\Longrightarrow a=0=b[/tex]
 
  • #13
HallsofIvy
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Yes, that is why you can't "prove [itex]e^x[/itex] and [itex]e^{2x}[/itex] are dependent": they are independent.
 
  • #14
matt grime
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Remember this is equal to the zero *function* in the vector space of real valued functions. The zero function is the one that is zero for all its inputs.
 
  • #15
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Is not the definition for linear indepedence the following:

for all a,b and x ,real Nos and [tex]ae^x + be^{2x}=0\Longrightarrow a=0=b[/tex]
Yes, that is why you can't "prove [itex]e^x[/itex] and [itex]e^{2x}[/itex] are dependent": they are independent.


Then the negation of the above definition should imply for linear dependency

But the negation of the above definition is:

There exist a.b, and ,x such that [tex]ae^x + be^{2x} = 0[/tex] and[tex]a\neq 0[/tex] and [tex]b\neq 0[/tex].

HENCE if we put a=2 ,b=-2 ,x=0 ,we satisfy the linear dependency definition
 
  • #16
matt grime
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That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

the negation of which is

(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.
 
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  • #17
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That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

.
Can you put that into a quantifier form??
 
  • #18
matt grime
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In what way is that not in a useful quantifier form (for those who like things like that)? Really, we shouldn't even bother with the quantifier "for all a,b", and putting things in the full on formal abstract quantifier notation just makes things far more opaque than they need to be.

You're making a very easy question seem very hard: can you find real numbers a and b not both equal to 0 so that

ae^x + be^2x

is the zero function? No - it has been proven several times in this thread.
 
  • #19
HallsofIvy
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The point is that you are working in a vector space of functions. When we say "af(x)+ bg(x)= 0" we mean it is equal to the 0 function- 0 for all x.
 
  • #20
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That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

the negation of which is

(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.
The negation of :

(for all a,b)(ae^x + be^2x =0for all x ===.> a=b=0) is

(there exist a,b) [~(ae^x + be^2x = 0 for all x ====> a=b=0)] ======>

(there exist a,b) [ ae^x = be^2x =0 there exist x AND a=/=0 ,b=/=0]


Whether you put the quantifier infront ,like i did, or at the end ,like you did we still have the same result.

The negation of [tex]\forall xPx[/tex] is : [tex]\exists x\neg Px[/tex] .

In our case Px is ( ae^x + be^2x =0 ======> a=b=0) and the negation of Px is:


ae^x = be^2x =0 and a=/=0,b=/=0

When you want to show that the function ae^x +be^2 is zero for all xεR ,YOU write :

for all xεR, ae^x + be^2x =0 OR ae^x + be^2x = 0, for all x.

To say that: for all x, is not part of the definition ,then what is part of??
 
  • #21
HallsofIvy
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The negation of :

(for all a,b)(ae^x + be^2x =0for all x ===.> a=b=0) is

(there exist a,b) [~(ae^x + be^2x = 0 for all x ====> a=b=0)] ======>

(there exist a,b) [ ae^x = be^2x =0 there exist x AND a=/=0 ,b=/=0]


Whether you put the quantifier infront ,like i did, or at the end ,like you did we still have the same result.

The negation of [tex]\forall xPx[/tex] is : [tex]\exists x\neg Px[/tex] .

In our case Px is ( ae^x + be^2x =0 ======> a=b=0) and the negation of Px is:


ae^x = be^2x =0 and a=/=0,b=/=0

When you want to show that the function ae^x +be^2 is zero for all xεR ,YOU write :

for all xεR, ae^x + be^2x =0 OR ae^x + be^2x = 0, for all x.

To say that: for all x, is not part of the definition ,then what is part of??
You did not quote all of what he said. He said "The for all x is not part of the definition of linear (in)dependence" and then followed with "it is part of the definition of what it means for the function to be zero" and that was the point of my last response:

We are talking about functions of x. Saying that "[itex]ae^x+ be^{-x}= 0[/itex]" means that the function [itex]f(x)= ae^x+ be^{-x}[/itex] is equal to the "0 function": g(x)= 0.
 
  • #22
matt grime
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To add to what Halls said: you have not take the negation of A => B correctly. The positioning of the quantifiers is very important: the negation of A => B is A and not(B), so the "for all" in there is not changed. You have negated A=>B and gotten, well, goodness knows what in relation to A and B.
 
  • #23
if [tex](e^x, e^{2x})[/tex] is linear denpendent, then [tex]e^{2x}=ke^{x},k\in R, k\ is\ constant \longrightarrow e^x=k[/tex],but [tex]e^x[/tex] is not a constant, so [tex](e^x, e^{2x})[/tex] is linear independent
 
  • #24
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You did not quote all of what he said. He said "The for all x is not part of the definition of linear (in)dependence" and then followed with "it is part of the definition of what it means for the function to be zero" and that was the point of my last response:

We are talking about functions of x. Saying that "[itex]ae^x+ be^{-x}= 0[/itex]" means that the function [itex]f(x)= ae^x+ be^{-x}[/itex] is equal to the "0 function": g(x)= 0.
What is your definition of linear Independence ,in symbolical form or not ??
 
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  • #25
HallsofIvy
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Sigh: A set of vectors
[tex]\{v_1, v_2, \cdot\cdot\cdot, v_n\}[/tex]
is independent
If the only set of scalars
[tex]\{a_1, a_2, \cdot\cdot\cdot, a_n\}[/tex]
such that
[tex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0[/tex]
is
[tex]a_1= a_2= \cdot\cdot\cdot= a_n= 0[/itex]

But the point you seem to be having trouble with is this: Since the left side of the first equation is a linear combination of vectors, the "0" on the right side is the 0 vector!

When we talk about functions being "independent" or "dependent" we are talking about the functions as members of some vector space- the set of all polynomials, continuous functions, differentiable functions, infinitely differentiable functions, etc. and the 0 vector is the 0 function- i.e. the function f such that f(x)= 0 for all x.

Given almost any set of functions, you can always find numbers such that the linear combination is 0 at one value of x. That has nothing to do with them being "independent".
 
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